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Author Topic: The Elimination Game  (Read 409 times)
Axeman
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« on: Dec 04, 2009 at 15:25 »

If You don't know how this works, Ya just add 1 for your favorite player listed and subtract 1 from your least Favorite Player. now I know we are all Steeler fans and we should like everyone on the team but this is a fun game just to see who would be the favorite. when someone reaches 0 they are out.

B. Rothlisberger 6 +1
H. Ward 5
R. Mendenhall 5
S. Holmes 5
H. Miller 5
J. Harrison 5
T. Polamalu 5
A. Smith 5
R. Clark 4 -1
L. Woodley 5


make sure you copy and paste from the person above you.
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If it doesn't matter who wins the game, then why keep score.
bamf16
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« Reply #1 on: Dec 04, 2009 at 15:41 »

If You don't know how this works, Ya just add 1 for your favorite player listed and subtract 1 from your least Favorite Player. now I know we are all Steeler fans and we should like everyone on the team but this is a fun game just to see who would be the favorite. when someone reaches 0 they are out.

B. Roethlisberger 6 +1
H. Ward 5
R. Mendenhall 5
S. Holmes 5
H. Miller 5
J. Harrison 5
T. Polamalu 5
A. Smith 5
R. Clark 4 -1
L. Woodley 5


make sure you copy and paste from the person above you.




L. Woodley + 1 (6)

S. Holmes - 1 (4)

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No one wants to hear about the labor pains, they just want to see the baby.
--Lou Brock
Finnegans Wake
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« Reply #2 on: Dec 04, 2009 at 15:51 »

Here's a little game I came up with.  Take all the players on the 53-man roster.  For each player, take his jersey number and divide it by the numerological value of his name.  For example, Hines Ward would be 96 divided by (H=8 + I=9 + N=14 + E=5 + S=19 + W=23 + A=1 + D=4).  For Hines, that comes to 86/101, or 0.8514851485148514... .

Now, sort all the player values from highest to lowest, and multiply the highest by the lowest, the second highest by the second lowest, etc.  There will be one odd, unmatched player at the median; square that player's value.  Let's say one such pairing might be Sepulveda-Kirschke, with a value of 1.0476539201.

Now, re-sort all the two-player pairings (and the odd man squared) and re-sort from high value to low.  You will have 27 such pairings.  Again, multiply highest by lowest, second highest by second lowest, and square the odd leftover.  You will have "four-man teams" with some odd squares thrown in, with 14 such values.

As you keep going, you will find that pairing off the 14 teams will have no odd man out, and can be reduced to 7 teams; this can be reduced to 4 teams (with an odd man); this can be reduced to 2 teams, then, which brings us to the final part of the game.

When you have paired up all the 53-man roster into two numerologically-and-jersey-number balanced squads, see which team has the greater numerological value.  Sort each team into starters at each actual football position.  Then, create a narrative explaining why the numerologically superior team should be better as a football team than the numerologically inferior team.

OK.  It's a game.  I didn't say it was a fun game, or a particularly interesting game, but it is a game, and one that will keep idle hands busy for a while...
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Out of my mind on Saturday night...
kluisi61
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« Reply #3 on: Dec 24, 2009 at 09:08 »

Here's a little game I came up with.  Take all the players on the 53-man roster.  For each player, take his jersey number and divide it by the numerological value of his name.  For example, Hines Ward would be 96 divided by (H=8 + I=9 + N=14 + E=5 + S=19 + W=23 + A=1 + D=4).  For Hines, that comes to 86/101, or 0.8514851485148514... .

Now, sort all the player values from highest to lowest, and multiply the highest by the lowest, the second highest by the second lowest, etc.  There will be one odd, unmatched player at the median; square that player's value.  Let's say one such pairing might be Sepulveda-Kirschke, with a value of 1.0476539201.

Now, re-sort all the two-player pairings (and the odd man squared) and re-sort from high value to low.  You will have 27 such pairings.  Again, multiply highest by lowest, second highest by second lowest, and square the odd leftover.  You will have "four-man teams" with some odd squares thrown in, with 14 such values.

As you keep going, you will find that pairing off the 14 teams will have no odd man out, and can be reduced to 7 teams; this can be reduced to 4 teams (with an odd man); this can be reduced to 2 teams, then, which brings us to the final part of the game.

When you have paired up all the 53-man roster into two numerologically-and-jersey-number balanced squads, see which team has the greater numerological value.  Sort each team into starters at each actual football position.  Then, create a narrative explaining why the numerologically superior team should be better as a football team than the numerologically inferior team.

OK.  It's a game.  I didn't say it was a fun game, or a particularly interesting game, but it is a game, and one that will keep idle hands busy for a while...

I really wish I had the time to write up a spreadsheet to do all of that. Wink
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OK Chris. Now that we have practiced kissing and cuddling, we'll practice eating out...at a fancy restaurant.

 - Lois Griffin

-----------------

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Fallen Sword - MMORPG that is very fun. No ads and it's free.

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